Given an array A of positive integers, call a (contiguous, not necessarily distinct) subarray of A good if the number of different integers in that subarray is exactly K.
(For example, [1,2,3,1,2] has 3 different integers: 1, 2, and 3.)
Return the number of good subarrays of A.
Example 1:
Input: A = [1,2,1,2,3], K = 2
Output: 7
Explanation: Subarrays formed with exactly 2 different integers: [1,2], [2,1], [1,2], [2,3], [1,2,1], [2,1,2], [1,2,1,2].
Example 2:
Input: A = [1,2,1,3,4], K = 3
Output: 3
Explanation: Subarrays formed with exactly 3 different integers: [1,2,1,3], [2,1,3], [1,3,4].
Note:
1 <= A.length <= 20000
1 <= A[i] <= A.length
1 <= K <= A.length
Solution:
class Solution {
public int subarraysWithKDistinct(int[] A, int K) {
int n = A.length;
int left = 0, right = 0;
int res = 0;
Map<Integer, Integer> map = new HashMap();
while (right < n) {
int in = A[right];
map.put(in, map.getOrDefault(in, 0) + 1);
if (map.size() == K) {
if (right == n - 1 || !map.containsKey(A[right + 1])) {
while (map.size() == K) {
// System.out.println(map);
res ++;
int out = A[left];
map.put(out, map.get(out) - 1);
if (map.get(out) == 0) map.remove(out);
left ++;
}
} else {
int tempLeft = left;
Map<Integer, Integer> tempMap = new HashMap();
tempMap.putAll(map);
while (tempMap.size() == K) {
// System.out.println(tempMap);
res ++;
int out = A[tempLeft];
tempMap.put(out, tempMap.get(out) - 1);
if (tempMap.get(out) == 0) tempMap.remove(out);
tempLeft ++;
}
}
}
right ++;
}
return res;
}
}